Modules of Finite Length - Cohn, page 61

[Math Amateur]

I am reading "Introduction to Ring Theory" by P. M. Cohn (Springer Undergraduate Mathematics Series)

In Chapter 2: Linear Algebras and Artinian Rings, on Page 61 we find a definition of the length of a module. Some analysis follows, as does a statement of Theorem 2.5. I need help to understand both the analysis and the theorem.

The relevant text on page 61 is as follows:View attachment 3287In the above text, we read:

" ... ... Now for any homomorphism\(\displaystyle f \ : \ M \to N \), we have the the isomorphisms:

\(\displaystyle M/ \text{ker } f \cong \text{im } f\)

\(\displaystyle \text{coker } f = N / \text{im f} \)

It follows that:

\(\displaystyle \ell (M) - \ell ( \text{ker } f ) = \ell ( \text{im f} )\)

and

\(\displaystyle \ell ( \text{coker } f ) = \mathcal{l} (N) - \ell ( \text{im f} ) \)

and so we obtain the equation

\(\displaystyle \ell (M) - \ell ( \text{ker } f ) = \ell (N) - \ell ( \text{coker } f ) \)

whenever all terms are defined. ... ... "

I need help to understand how the isomorphisms

\(\displaystyle M/ \text{ker } f \cong \text{im } f \)

\(\displaystyle \text{coker } f = N / \text{im f} \)

imply that

\(\displaystyle \ell (M) - \ell ( \text{ker } f ) = \ell ( \text{im f} )\)

and

\(\displaystyle \ell ( \text{coker } f ) = \ell (N) - \ell ( \text{im f} ) \)

and then, further, how Theorem 2.5 follows.Would appreciate some help ...

Peter***EDIT*** Thanks to Mark for his help with the latex code!

 

[mathbalarka]

This is a special case of a more general fact. Let there be a short exact sequence of modules

$$0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$$

Where $f$ is the injective homomorphism $A \to B$ and $g$ is the surjective homomorphism $B \to C$. We claim that $\ell(C) = \ell(B) - \ell(A)$.

$\{0\} = C_0 \subset C_1 \subset \cdots \subset C_n = C$ be a composition series for $C$ of length $\ell(C) = n$ and

$\{0\} = A_0 \subset A_1 \subset \cdots \subset A_m = A$ be composition series for $A$ of length $\ell(A) = m$.

However, as $f$ and $g^{-1}$ are injective, we have the composition series

$\color{blue}{\{0\} = f(A_0) \subset f(A_1) \subset \cdots \subset f(A_m) = f(A)}$ for $\text{im} \, f$ and the composition series

$\color{red}{\{\text{ker} \, g\} = g^{-1}(C_0)/\text{ker} \, g \subset g^{-1}(C_1)/\text{ker}\, g \subset \cdots \subset g^{-1}(C_n = C)/\text{ker} \, g = B/\text{ker} \, g}$ for $B/\text{ker} \, g$

Combining the last two series one gets a chain of submodules

$\{0\} = f(A_0) \subset f(A_1) \subset \cdots \subset f(A_m\!=\!A) = \text{im} \, f = \text{ker} \, g = g^{-1}(C_0) \subset g^{-1}(C_1) \subset \cdots \subset g^{-1}(C_n \! = \! C) = B$

Following the third isomorphism theorem for modules, $g^{-1}(C_k)/g^{-1}(C_{k-1}) \cong \left ( g^{-1}(C_k)/\text{ker}\, g \right) / \left( g^{-1}(C_{k-1})/\text{ker}\, g \right)$ is a simple module for $1 \leq k \leq n$ and the rest of the composition factors are automatically simple as they are the same factors which appear in the series for $f(A)$. Thus, we have a composition series for $B$ of length $m + n = \ell(A) + \ell(C)$.

It follows that $\ell(\text{im}\, f) = \ell(M) - \ell(\text{ker}\, f)$ and $\ell(\text{coker}\, f) = \ell(N) - \ell(\text{im}\, f)$ by noting that $0 \to \text{ker} \, f \to M \to \text{im} \, f \to 0$ and $0 \to \text{im} \, f \to \text{N} \to \text{coker} f \to 0$ are short exact sequence, the former a consequence of first isomorphism theorem and the latter following from definition.

 

[Math Amateur]

This is a special case of a more general fact. Let there be a short exact sequence of modules

$$0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$$

Where $f$ is the injective homomorphism $A \to B$ and $g$ is the surjective homomorphism $B \to C$. We claim that $\ell(C) = \ell(B) - \ell(A)$.

$\{0\} = C_0 \subset C_1 \subset \cdots \subset C_n = C$ be a composition series for $C$ of length $\ell(C) = n$ and

$\{0\} = A_0 \subset A_1 \subset \cdots \subset A_m = A$ be composition series for $A$ of length $\ell(A) = m$.

However, as $f$ and $g^{-1}$ are injective, we have the composition series

$\color{blue}{\{0\} = f(A_0) \subset f(A_1) \subset \cdots \subset f(A_m) = f(A)}$ for $\text{im} \, f$ and the composition series

$\color{red}{\{\text{ker} \, g\} = g^{-1}(C_0)/\text{ker} \, g \subset g^{-1}(C_1)/\text{ker}\, g \subset \cdots \subset g^{-1}(C_n = C)/\text{ker} \, g = B/\text{ker} \, g}$ for $B/\text{ker} \, g$

Combining the last two series one gets a chain of submodules

$\{0\} = f(A_0) \subset f(A_1) \subset \cdots \subset f(A_m\!=\!A) = \text{im} \, f = \text{ker} \, g = g^{-1}(C_0) \subset g^{-1}(C_1) \subset \cdots \subset g^{-1}(C_n \! = \! C) = B$

Following the third isomorphism theorem for modules, $g^{-1}(C_k)/g^{-1}(C_{k-1}) \cong \left ( g^{-1}(C_k)/\text{ker}\, g \right) / \left( g^{-1}(C_{k-1})/\text{ker}\, g \right)$ is a simple module for $1 \leq k \leq n$ and the rest of the composition factors are automatically simple as they are the same factors which appear in the series for $f(A)$. Thus, we have a composition series for $B$ of length $m + n = \ell(A) + \ell(C)$.

It follows that $\ell(\text{im}\, f) = \ell(M) - \ell(\text{ker}\, f)$ and $\ell(\text{coker}\, f) = \ell(N) - \ell(\text{im}\, f)$ by noting that $0 \to \text{ker} \, f \to M \to \text{im} \, f \to 0$ and $0 \to \text{im} \, f \to \text{N} \to \text{coker} f \to 0$ are short exact sequence, the former a consequence of first isomorphism theorem and the latter following from definition.

Thanks for the help Mathbalarka ... but I need some help with this ...

Can you help me with the following questions ... ...



Question 1

Given the exact sequence

$$0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$$

how do we know that the composition series

$\{0\} = C_0 \subset C_1 \subset \cdots \subset C_n = C$

and

$\{0\} = A_0 \subset A_1 \subset \cdots \subset A_m = A$

actually exist?

Question 2

Given that for any function f we have that

\(\displaystyle A_i \subset A_{i+1} \Longrightarrow f( A_i ) \subset f( A_{i+1} ) \)

why do we need f to be injective to claim that

$\{0\} = A_0 \subset A_1 \subset \cdots \subset A_m = A$

Question 3

Can you help me to understand exactly how$\color{red}{\{\text{ker} \, g\} = g^{-1}(C_0)/\text{ker} \, g \subset g^{-1}(C_1)/\text{ker}\, g \subset \cdots \subset g^{-1}(C_n = C)/\text{ker} \, g = B/\text{ker} \, g}$ for $B/\text{ker} \, g$

follows from the assumptions made.

Hope you can help.

Peter

 

[mathbalarka]

Question 1
Given the exact sequence

$$0 \to A \stackrel{f}{\to} B \stackrel{g}{\to} C \to 0$$

how do we know that the composition series

$\{0\} = C_0 \subset C_1 \subset \cdots \subset C_n = C$

and

$\{0\} = A_0 \subset A_1 \subset \cdots \subset A_m = A$

actually exist?

It is absolutely NOT nessesary that composition series for $A$ and $C$ exist. We are ASSUMING that they exists, i.e., that $\ell(A)$ and $\ell(C)$ are finite. In fact, $\ell(B)$ is finite if and only if $\ell(A)$ and $\ell(C)$ are (the "if" part we already proved, what about the "only if" part?).

Question 2

Given that for any function f we have that

\(\displaystyle A_i \subset A_{i+1} \Longrightarrow f( A_i ) \subset f( A_{i+1} ) \)

Certainly the implication doesn't hold for arbitrary maps $f$. For example, take the zero map $f : x \mapsto 0$, in which case $f(D) = \{0\}$ for any subset $D$ of the domain. Hence, $f(A_i) = f(A_{i+1}) = 0$, thus strict inclusion doesn't hold. Your claim is false.

Question 3

Can you help me to understand exactly how$\color{red}{\{\text{ker} \, g\} = g^{-1}(C_0)/\text{ker} \, g \subset g^{-1}(C_1)/\text{ker}\, g \subset \cdots \subset g^{-1}(C_n = C)/\text{ker} \, g = B/\text{ker} \, g}$ for $B/\text{ker} \, g$

Note that the inclusions are a consequence of injectivity of $g^{-1}$. Can you prove that the composition factors are simple?

 

[Math Amateur]

It is absolutely NOT nessesary that composition series for $A$ and $C$ exist. We are ASSUMING that they exists, i.e., that $\ell(A)$ and $\ell(C)$ are finite. In fact, $\ell(B)$ is finite if and only if $\ell(A)$ and $\ell(C)$ are (the "if" part we already proved, what about the "only if" part?).

Certainly the implication doesn't hold for arbitrary maps $f$. For example, take the zero map $f : x \mapsto 0$, in which case $f(D) = \{0\}$ for any subset $D$ of the domain. Hence, $f(A_i) = f(A_{i+1}) = 0$, thus strict inclusion doesn't hold. Your claim is false.

Note that the inclusions are a consequence of injectivity of $g^{-1}$. Can you prove that the composition factors are simple?

Sorry to be slow Mathbalarka ... but I need some more help with basics here ...So my (simple) questions are as follows1) From the definition of an exact sequence we have that \(\displaystyle g\) is a surjective homomorphism ... how does it follow from this that \(\displaystyle g^{-1}\) is injective ... can you give a simple proof?

(EDIT: Doesn't \(\displaystyle g\) have to be a bijection for \(\displaystyle g^{-1}\) to exist - mind you, if that were the case, then \(\displaystyle g^{-1}\) would also be a bijection and hence certainly injective)
2) Given that \(\displaystyle g^{-1}\) is injective, how exactly does it follow that $\color{red}{\{\text{ker} \, g\} = g^{-1}(C_0)/\text{ker} \, g \subset g^{-1}(C_1)/\text{ker}\, g \subset \cdots \subset g^{-1}(C_n = C)/\text{ker} \, g = B/\text{ker} \, g}$ for $B/\text{ker} \, g$

Can you please give a simple explicit proof?
Would be grateful for you help in these matters.

Peter

 

[Math Amateur]

Sorry to be slow Mathbalarka ... but I need some more help with basics here ...So my (simple) questions are as follows1) From the definition of an exact sequence we have that \(\displaystyle g\) is a surjective homomorphism ... how does it follow from this that \(\displaystyle g^{-1}\) is injective ... can you give a simple proof?

(EDIT: Doesn't \(\displaystyle g\) have to be a bijection for \(\displaystyle g^{-1}\) to exist - mind you, if that were the case, then \(\displaystyle g^{-1}\) would also be a bijection and hence certainly injective)
2) Given that \(\displaystyle g^{-1}\) is injective, how exactly does it follow that $\color{red}{\{\text{ker} \, g\} = g^{-1}(C_0)/\text{ker} \, g \subset g^{-1}(C_1)/\text{ker}\, g \subset \cdots \subset g^{-1}(C_n = C)/\text{ker} \, g = B/\text{ker} \, g}$ for $B/\text{ker} \, g$

Can you please give a simple explicit proof?
Would be grateful for you help in these matters.

Peter

I must say I am still struggling with the most basic elements of my original question ...

I would welcome a detailed, simple and direct (but formal and explicit) demonstration that

\(\displaystyle M/ \text{ker } f \cong \text{im } f \)

implies that

\(\displaystyle \ell (M) - \ell ( \text{ker } f ) = \ell ( \text{im f} )\)

Hope someone can provide such a demonstration ... ...

Peter

 

[mathbalarka]

I owe an apology to Peter for not answering his doubts. I am at the moment very busy with some works so haven't been able to draw some time to answer those.

Just to let everyone know, if someone wants to answer Peter's questions and resolve the doubts, he/she is most welcome. If I find after finishing the works that nobody has answered, I will definitely have a look.

 

[Math Amateur]

I owe an apology to Peter for not answering his doubts. I am at the moment very busy with some works so haven't been able to draw some time to answer those.

Just to let everyone know, if someone wants to answer Peter's questions and resolve the doubts, he/she is most welcome. If I find after finishing the works that nobody has answered, I will definitely have a look.

Hi Mathbalarka ... Just a note to say you have helped me considerably lately and I am most grateful for your insights and help ...

So ... there is absolutely no need to apologize ... indeed I am sorry I haven't followed you completely (yet ... I am still reflecting on what you have said ... ) ... I was just indicating that I was still struggling and I wanted to limit the question a bit ...

Thanks again for all your insightful help!

Peter

 

[Euge]

I must say I am still struggling with the most basic elements of my original question ...

I would welcome a detailed, simple and direct (but formal and explicit) demonstration that

\(\displaystyle M/ \text{ker } f \cong \text{im } f \)

implies that

\(\displaystyle \ell (M) - \ell ( \text{ker } f ) = \ell ( \text{im f} )\)

Hope someone can provide such a demonstration ... ...

Peter

Hi Peter,

Since $M/\text{ker }f \cong \text{im}f$, $\ell(M/\text{ker }f) = \ell(\text{im }f)$. So it suffices to show that $\ell(M/\text{ker }f) = \ell(M) - \ell(\text{ker }f)$. To do this, suppose $M/\text{ker}f$ has length $r$ and $\text{ker }f$ has length $s$. The quotient $M/\text{ker }f$ has a series

\(\displaystyle \text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f\)

such that $A_i/A_{i-1}$ is simple for $i = 1, 2,\ldots, r$. For each $i$, let $B_i$ be the preimage of $A_i$ under the natural projection $\pi : M \to M/\text{ker }f$. The series above ascends to the series

\(\displaystyle \text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M\)

For $i = 1, 2,\ldots r$, $B_i/B_{i-1}$ can be naturally identified with $A_i/A_{i-1}$ by an isomorphism induced by the map $B_i \to A_i/A_{i-1}$, $x \mapsto \pi(x) + A_{i-1}$. Therefore each quotient $B_i/B_{i-1}$ is simple.

Since $\text{ker }f$ has length $s$, it has a series

\(\displaystyle 0 = C_0 \subset C_1 \subset \cdots \subset C_s = \text{ker } f\)

such that $C_i/C_{i-1}$ is simple for $i = 1, 2, \ldots, r$. Now we have a composition series

\(\displaystyle 0 = C_0 \subset C_1 \subset \cdots \subset C_s = A_0 \subset A_1 \subset \cdots \subset A_r = M\)

which shows that $M$ has length $r + s$.

 

[Math Amateur]

Hi Peter,

Since $M/\text{ker }f \cong \text{im}f$, $\ell(M/\text{ker }f) = \ell(\text{im }f)$. So it suffices to show that $\ell(M/\text{ker }f) = \ell(M) - \ell(\text{ker }f)$. To do this, suppose $M/\text{ker}f$ has length $r$ and $\text{ker }f$ has length $s$. The quotient $M/\text{ker }f$ has a series

\(\displaystyle \text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f\)

such that $A_i/A_{i-1}$ is simple for $i = 1, 2,\ldots, r$. For each $i$, let $B_i$ be the preimage of $A_i$ under the natural projection $\pi : M \to M/\text{ker }f$. The series above ascends to the series

\(\displaystyle \text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M\)

For $i = 1, 2,\ldots r$, $B_i/B_{i-1}$ can be naturally identified with $A_i/A_{i-1}$ by an isomorphism induced by the map $B_i \to A_i/A_{i-1}$, $x \mapsto \pi(x) + A_{i-1}$. Therefore each quotient $B_i/B_{i-1}$ is simple.

Since $\text{ker }f$ has length $s$, it has a series

\(\displaystyle 0 = C_0 \subset C_1 \subset \cdots \subset C_s = \text{ker } f\)

such that $C_i/C_{i-1}$ is simple for $i = 1, 2, \ldots, r$. Now we have a composition series

\(\displaystyle 0 = C_0 \subset C_1 \subset \cdots \subset C_s = A_0 \subset A_1 \subset \cdots \subset A_r = M\)

which shows that $M$ has length $r + s$.

Thanks for the help Euge ... but just a few simple clarifications/questions ...

Question 1

You write:

" ... ...Since $M/\text{ker }f \cong \text{im}f$, $\ell(M/\text{ker }f) = \ell(\text{im }f)$. ... ... "

I know that this seems highly intuitive but can you give an explicit demonstration of why this is true?
Question 2

You write: " ... ... the quotient $M/\text{ker }f$ has a series

\(\displaystyle \text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f\) ... ..."

Why must this series start with \text{ker }f? ... ... Why, for example, can't the series start with a submodule A_0 \subset \text{ker }f?

Question 3

You write:" ... ... The series above ascends to the series

\(\displaystyle \text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M\) ... ... "What do you mean by the series "ascends" to the series ... ?

Hope you can clarify/help ...

Peter

 

[Euge]

Thanks for the help Euge ... but just a few simple clarifications/questions ...

Question 1

You write:

" ... ...Since $M/\text{ker }f \cong \text{im}f$, $\ell(M/\text{ker }f) = \ell(\text{im }f)$. ... ... "

I know that this seems highly intuitive but can you give an explicit demonstration of why this is true?

Suppose $A$ and $B$ are isomorphic modules such that $A$ has length $r$. I'll show that $B$ also has length $r$. Since $A$ has length $r$, it admits a composition series

\(\displaystyle 0 = A_0 \subset A_1 \subset \cdots \subset A_r = A.\)

Let $\phi : A \to B$ be an isomorphism of $A$ onto $B$. Define $B_i = \phi(A_i)$ for $i = 0, 1, \ldots r$. Then $B_0 = 0$, $B_{i-1} \subset B_i$ for $i = 1, 2,\ldots, r$, and $B_r = B$. Thus

\(\displaystyle 0 = B_0 \subset B_1 \subset \cdots \subset B_r = B.\)

Since $\phi$ is an isomorphism, there is, for each $i \in \{1,2,\ldots, r\}$, isomorphisms $B_i/B_{i-1} \cong A_i/A_{i-1}$ induced by the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$. Therefore, as each factor module $A_i/A_{i-1}$ is simple, so are the factor modules $B_i/B_{i-1}$. So the series for $B$ is a composition series with $B_r = B$. Hence, the length of $B$ is $r$.

Question 2

You write: " ... ... the quotient $M/\text{ker }f$ has a series

\(\displaystyle \text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f\) ... ..."

Why must this series start with \text{ker }f? ... ... Why, for example, can't the series start with a submodule A_0 \subset \text{ker }f?

A composition series starts with zero, and $\text{ker} f$ is the zero of $M/\text{ker} f$.

Question 3

You write:" ... ... The series above ascends to the series

\(\displaystyle \text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M\) ... ... "What do you mean by the series "ascends" to the series ... ?

I'm just saying that the series $\text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f$ for $M/\text{ker } f$ induces the series $\text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M$ for $M$.

 

[Math Amateur]

Suppose $A$ and $B$ are isomorphic modules such that $A$ has length $r$. I'll show that $B$ also has length $r$. Since $A$ has length $r$, it admits a composition series

\(\displaystyle 0 = A_0 \subset A_1 \subset \cdots \subset A_r = A.\)

Let $\phi : A \to B$ be an isomorphism of $A$ onto $B$. Define $B_i = \phi(A_i)$ for $i = 0, 1, \ldots r$. Then $B_0 = 0$, $B_{i-1} \subset B_i$ for $i = 1, 2,\ldots, r$, and $B_r = B$. Thus

\(\displaystyle 0 = B_0 \subset B_1 \subset \cdots \subset B_r = B.\)

Since $\phi$ is an isomorphism, there is, for each $i \in \{1,2,\ldots, r\}$, isomorphisms $B_i/B_{i-1} \cong A_i/A_{i-1}$ induced by the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$. Therefore, as each factor module $A_i/A_{i-1}$ is simple, so are the factor modules $B_i/B_{i-1}$. So the series for $B$ is a composition series with $B_r = B$. Hence, the length of $B$ is $r$.

A composition series starts with zero, and $\text{ker} f$ is the zero of $M/\text{ker} f$.
I'm just saying that the series $\text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f$ for $M/\text{ker } f$ induces the series $\text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M$ for $M$.

Thanks so much Euge ... just working through this post now ... appreciate your help ...

Peter

 

[Math Amateur]

Suppose $A$ and $B$ are isomorphic modules such that $A$ has length $r$. I'll show that $B$ also has length $r$. Since $A$ has length $r$, it admits a composition series

\(\displaystyle 0 = A_0 \subset A_1 \subset \cdots \subset A_r = A.\)

Let $\phi : A \to B$ be an isomorphism of $A$ onto $B$. Define $B_i = \phi(A_i)$ for $i = 0, 1, \ldots r$. Then $B_0 = 0$, $B_{i-1} \subset B_i$ for $i = 1, 2,\ldots, r$, and $B_r = B$. Thus

\(\displaystyle 0 = B_0 \subset B_1 \subset \cdots \subset B_r = B.\)

Since $\phi$ is an isomorphism, there is, for each $i \in \{1,2,\ldots, r\}$, isomorphisms $B_i/B_{i-1} \cong A_i/A_{i-1}$ induced by the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$. Therefore, as each factor module $A_i/A_{i-1}$ is simple, so are the factor modules $B_i/B_{i-1}$. So the series for $B$ is a composition series with $B_r = B$. Hence, the length of $B$ is $r$.

A composition series starts with zero, and $\text{ker} f$ is the zero of $M/\text{ker} f$.
I'm just saying that the series $\text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f$ for $M/\text{ker } f$ induces the series $\text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M$ for $M$.

Hi Euge,

Thanks again for your help ... but just a clarification ...

You write:

" ... ... A composition series starts with zero, and $\text{ker} f$ is the zero of $M/\text{ker} f$. ... ... "Yes, OK, get that ... but then you write:

" ... ... I'm just saying that the series $\text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f$ for $M/\text{ker } f$ induces the series $\text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M$ for $M$. ... ... "But in this case \(\displaystyle \text{ker }f\) is not \(\displaystyle \{ 0 \}\) but may be a proper, nontrivial subset of \(\displaystyle M\), so can't we, in this case, have the composition series start with a submodule \(\displaystyle B_0 \subset \text{ker }f\)?

Can you help?

Peter

 

[Euge]

The series with $B_i$ is not a composition series.

 

[Math Amateur]

Suppose $A$ and $B$ are isomorphic modules such that $A$ has length $r$. I'll show that $B$ also has length $r$. Since $A$ has length $r$, it admits a composition series

\(\displaystyle 0 = A_0 \subset A_1 \subset \cdots \subset A_r = A.\)

Let $\phi : A \to B$ be an isomorphism of $A$ onto $B$. Define $B_i = \phi(A_i)$ for $i = 0, 1, \ldots r$. Then $B_0 = 0$, $B_{i-1} \subset B_i$ for $i = 1, 2,\ldots, r$, and $B_r = B$. Thus

\(\displaystyle 0 = B_0 \subset B_1 \subset \cdots \subset B_r = B.\)

Since $\phi$ is an isomorphism, there is, for each $i \in \{1,2,\ldots, r\}$, isomorphisms $B_i/B_{i-1} \cong A_i/A_{i-1}$ induced by the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$. Therefore, as each factor module $A_i/A_{i-1}$ is simple, so are the factor modules $B_i/B_{i-1}$. So the series for $B$ is a composition series with $B_r = B$. Hence, the length of $B$ is $r$.

A composition series starts with zero, and $\text{ker} f$ is the zero of $M/\text{ker} f$.
I'm just saying that the series $\text{ker }f = A_0 \subset A_1 \subset \cdots \subset A_r = M/\text{ker }f$ for $M/\text{ker } f$ induces the series $\text{ker }f = B_0 \subset B_1 \subset \cdots \subset B_r = M$ for $M$.

So sorry to be slow here, Euge ... but can you help a bit further ... ...

In your post above you write:

" ... ... Suppose $A$ and $B$ are isomorphic modules such that $A$ has length $r$. I'll show that $B$ also has length $r$. Since $A$ has length $r$, it admits a composition series

\(\displaystyle 0 = A_0 \subset A_1 \subset \cdots \subset A_r = A.\)

Let $\phi : A \to B$ be an isomorphism of $A$ onto $B$. Define $B_i = \phi(A_i)$ for $i = 0, 1, \ldots r$. Then $B_0 = 0$, $B_{i-1} \subset B_i$ for $i = 1, 2,\ldots, r$, and $B_r = B$. Thus

\(\displaystyle 0 = B_0 \subset B_1 \subset \cdots \subset B_r = B.\)

Since $\phi$ is an isomorphism, there is, for each $i \in \{1,2,\ldots, r\}$, isomorphisms $B_i/B_{i-1} \cong A_i/A_{i-1}$ induced by the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$. Therefore, as each factor module $A_i/A_{i-1}$ is simple, so are the factor modules $B_i/B_{i-1}$. So the series for $B$ is a composition series with $B_r = B$. Hence, the length of $B$ is $r$. ... ... "

In particular in the above I am studying and reflecting upon your statement:

" ... ... Since $\phi$ is an isomorphism, there is, for each $i \in \{1,2,\ldots, r\}$, isomorphisms $B_i/B_{i-1} \cong A_i/A_{i-1}$ induced by the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$. ... ... "

Can you explain more explicitly how you get $B_i/B_{i-1} \cong A_i/A_{i-1}$? How exactly is this the case? How/why exactly does it follow from the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$?

I hope you can help ... ...

Apologies again for being a bit slow ...

Peter

 

[Math Amateur]

The series with $B_i$ is not a composition series.

Oh, indeed you are right, of course ...

Thanks,

Peter

 

[Euge]

So sorry to be slow here, Euge ... but can you help a bit further ... ...

In your post above you write:

" ... ... Suppose $A$ and $B$ are isomorphic modules such that $A$ has length $r$. I'll show that $B$ also has length $r$. Since $A$ has length $r$, it admits a composition series

\(\displaystyle 0 = A_0 \subset A_1 \subset \cdots \subset A_r = A.\)

Let $\phi : A \to B$ be an isomorphism of $A$ onto $B$. Define $B_i = \phi(A_i)$ for $i = 0, 1, \ldots r$. Then $B_0 = 0$, $B_{i-1} \subset B_i$ for $i = 1, 2,\ldots, r$, and $B_r = B$. Thus

\(\displaystyle 0 = B_0 \subset B_1 \subset \cdots \subset B_r = B.\)

Since $\phi$ is an isomorphism, there is, for each $i \in \{1,2,\ldots, r\}$, isomorphisms $B_i/B_{i-1} \cong A_i/A_{i-1}$ induced by the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$. Therefore, as each factor module $A_i/A_{i-1}$ is simple, so are the factor modules $B_i/B_{i-1}$. So the series for $B$ is a composition series with $B_r = B$. Hence, the length of $B$ is $r$. ... ... "

In particular in the above I am studying and reflecting upon your statement:

" ... ... Since $\phi$ is an isomorphism, there is, for each $i \in \{1,2,\ldots, r\}$, isomorphisms $B_i/B_{i-1} \cong A_i/A_{i-1}$ induced by the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$. ... ... "

Can you explain more explicitly how you get $B_i/B_{i-1} \cong A_i/A_{i-1}$? How exactly is this the case? How/why exactly does it follow from the maps $A_i \to B_i/B_{i-1}$, $x \mapsto \phi(x) + B_{i-1}$?

I hope you can help ... ...

Apologies again for being a bit slow ...

Peter

Fix $i\in \{1,2,\ldots, r\}$. The epimorphism $A_i \mapsto B_i/B_{i-1}$, $x\mapsto \phi(x) + B_{i-1}$ has kernel $K$ consisting of all $x \in A_i$ such that $\phi(x) \in B_{i-1}$. Since $\phi$ is bijective and $B_{i-1} = \phi(A_{i-1})$, $\phi(x)\in B_{i-1}$ if and only if $x\in A_{i-1}$. Thus $K = A_{i-1}$ and the first isomorphism theorem for modules yields $A_i/A_{i-1} \cong B_i/B_{i-1}$.

 

[Math Amateur]

Fix $i\in \{1,2,\ldots, r\}$. The epimorphism $A_i \mapsto B_i/B_{i-1}$, $x\mapsto \phi(x) + B_{i-1}$ has kernel $K$ consisting of all $x \in A_i$ such that $\phi(x) \in B_{i-1}$. Since $\phi$ is bijective and $B_{i-1} = \phi(A_{i-1})$, $\phi(x)\in B_{i-1}$ if and only if $x\in A_{i-1}$. Thus $K = A_{i-1}$ and the first isomorphism theorem for modules yields $A_i/A_{i-1} \cong B_i/B_{i-1}$.

Appreciate your help, Euge ... ... just working through this post now ...

Peter

 

[Deveno]

Something that may, or may not be helpful, here.

When Euge says (in post #9): "...ascends to a series $B_0 \subset B_1 \subset \cdots \subset B_r$.."

he is talking about the same $g^{-1}$ that mathbalarka uses in post #2.

Let's be a bit more explicitly about this "pre-image map" (not an INVERSE map, as $g$ is not injective).

Say we have the map:

$g: M \to M/N$ given by $g(m) = m + N$.

What does a submodule of $M/N$ look like?

I claim it looks like this: $L/N$ for some submodule $L$ of $M$ containing $N$.

It should be clear that if $L$ is a submodule of $M$ containing $N$, that $L/N$ "makes sense", and certainly the set of cosets:

$\{l + N: l \in L\}$ is a submodule of $M/N$.

Just as clearly, this set of cosets ($L/N$) is $g(L)$.

What may NOT be obvious is that ANY submodule of $M/N$ is of this form.

So suppose we have a submodule, say $K$ of $M/N$. Define $L = g^{-1}(K)$.

As far as we know, this is just a set. Suppose $x,y \in L$. Since $g$ is a module homomorphism (and thus an abelian group homomorphism):

$g(x - y) = g(x) - g(y)$.

Since $x,y \in L$, it must be the case that $g(x),g(y) \in K$.

Thus $g(x) - g(y) \in K$ (since this must be true for any two elements of $K$-the "one step group test").

Therefore, $g(x-y) \in K$, so $x - y$ is a pre-image of an element of $K$, and is thus in $L$ (because that is what we defined $L$ to be).

Thus $L$ satisfies the "one step group test", and is thus a subgroup of $(M,+)$. Since $M$ is an abelian group under addition, $L$ is, as well.

So now all we need to do is check $L$ is closed under scalar multiplication. Let $a \in R$ (the ring our module is over).

We know that $g(a\cdot x) = a\cdot g(x)$ (the LHS is the ring action of $M$, the RHS is the ring action of $M/N$), because $g$ is a module homomorphism.

Now if $x \in L$, then $g(x) \in K$, and since $K$ is a submodule of $M/N$ it IS closed under the ring action.

So if $x \in L$, then $a\cdot g(x) \in K$, for any $a \in R$.

But as we saw above, this is $g(a\cdot x)$, so we see that $a \cdot x \in L$, whenever $x \in L$. Thus $L$ is a bona-fide submodule of $M$. But now:

$L/N = g(L) = g(g^{-1}(K)) = K$, so indeed any submodule of $M/N$ is of the form indicated above (since $K$ was chosen arbitrarily).

Is the submodule $L$ we describe above unique? Well, in general, no. But if we insist that $L$ contain $N$, then yes.

Suppose that $g(L) = g(L') = L/N$, and that $N \subseteq L,L'$.

Pick any $x' \in L'$. Then $g(x') = x' + N = l + N$, for some $l \in L$. It follows that $x' - l \in N$, so:

$g(x' - l) = N = 0 + N = 0_{M/N}$.

Since $g$ is a homomorphism, $g(x') = g(l)$. Recall how we defined $L$, as the pre-image under $g$ of $L/N$.

Since $x'$ is such a pre-image, $x' \in L$, and thus $L' \subseteq L$.

A similar argument with the roles of $L,L'$ reversed show that $L \subseteq L'$.

THIS is what mathbalarka means when he says $g^{-1}$ is injective. Given a composition series for $M/N$, we can "pull it back" to a "partial composition series" starting with $N$ and ending with $M$.

We just MERGE that with the composition series for $N$, to obtain a composition series for $M$.

*****************************

Here is an example:

Take the abelian group $\Bbb Z_{210}$. Let's find a composition series for it two different ways:

One such series is:

$\{0\} \subset \langle [105]_{210}\rangle \subset \langle [70]_{210},[105]_{210}\rangle \subset \langle [42]_{210},[70]_{210},[105]_{210}\rangle \subset \Bbb Z_{210}$

For example, we see that $\langle [70]_{210},[105]_{210}\rangle/\langle [105]_{210}\rangle$ has order 3 and so has no non-trivial submodules (in case you're wondering, the cosets are, explicitly:

$[0]_{210} + \langle [105]_{210}\rangle = \{[0]_{210},[105]_{210}\}$
$[35]_{210} + \langle [105]_{210}\rangle = \{[35]_{210},[140]_{210}\}$
$[70]_{210} + \langle [105]_{210}\rangle = \{[70]_{210},[175]_{210}\}$).

This composition series has length 4.

Now let's consider what we get choosing $N = \langle [70]_{210},[105]_{210}\rangle$.

A composition series for $N$ is clearly:

$\{[0]_{210}\} \subset \langle [105]_{210}\rangle \subset N$.

Instead of working with $\Bbb Z_{210}/N$ directly, let's work with the isomorphic group $\Bbb Z_{35}$, obtained from the surjective homomorphism:

$\phi: [k]_{210} \mapsto [k]_{35}$ (Verify that $N$ is indeed the kernel. $N$ is clearly contained in the kernel).

Now $\Bbb Z_{35}/\langle [7]_{35}\rangle$ has order 7, so we obtain the composition series for $\Bbb Z_{35}$ of:

$\{[0]_{35}\} \subset \langle [7]_{35}\rangle \subset \Bbb Z_{35}$.

"Pulling this back" to $\Bbb Z_{210}$, we have:

$\phi^{-1}(\{[0]_{35}\}) = N$
$\phi^{-1}(\Bbb Z_{35}) = \Bbb Z_{210}$

so the only thing of interest is determining $\phi^{-1}(\langle [7]_{35}\rangle)$. That is, we need to find the pre-images of:

$[0]_{35},[7]_{35},[14]_{35},[21]_{35},[28]_{35}$.

These are:

$[0]_{210},[35]_{210},[70]_{210},[105]_{210},[140]_{210},[175]_{210}$ for $[0]_{35}$
$[7]_{210},[42]_{210},[77]_{210},[112]_{210},[147]_{210},[182]_{210}$ for $[7]_{35}$, and so on (keep going up by 7's).

Note that this set of pre-images contains $[42]_{210},[70]_{210},[105]_{210}$, so it contains the submodule:

$\langle [42]_{210},[70]_{210},[105]_{210}\rangle$

Note as well, the submodule of the pre-images has order 30, so it is maximal in $\Bbb Z_{210}$ as 210/30 = 7 is prime. So the pre-image set here actually is the same submodule we used in the first composition series.

These two series have length 2 each, and lo and behold, 2 + 2 = 4.

 

[Math Amateur]

Something that may, or may not be helpful, here.

When Euge says (in post #9): "...ascends to a series $B_0 \subset B_1 \subset \cdots \subset B_r$.."

he is talking about the same $g^{-1}$ that mathbalarka uses in post #2.

Let's be a bit more explicitly about this "pre-image map" (not an INVERSE map, as $g$ is not injective).

Say we have the map:

$g: M \to M/N$ given by $g(m) = m + N$.

What does a submodule of $M/N$ look like?

I claim it looks like this: $L/N$ for some submodule $L$ of $M$ containing $N$.

It should be clear that if $L$ is a submodule of $M$ containing $N$, that $L/N$ "makes sense", and certainly the set of cosets:

$\{l + N: l \in L\}$ is a submodule of $M/N$.

Just as clearly, this set of cosets ($L/N$) is $g(L)$.

What may NOT be obvious is that ANY submodule of $M/N$ is of this form.

So suppose we have a submodule, say $K$ of $M/N$. Define $L = g^{-1}(K)$.

As far as we know, this is just a set. Suppose $x,y \in L$. Since $g$ is a module homomorphism (and thus an abelian group homomorphism):

$g(x - y) = g(x) - g(y)$.

Since $x,y \in L$, it must be the case that $g(x),g(y) \in K$.

Thus $g(x) - g(y) \in K$ (since this must be true for any two elements of $K$-the "one step group test").

Therefore, $g(x-y) \in L$, so $x - y$ is a pre-image of an element of $K$, and is thus in $L$ (because that is what we defined $L$ to be).

Thus $L$ satisfies the "one step group test", and is thus a subgroup of $(M,+)$. Since $M$ is an abelian group under addition, $L$ is, as well.

So now all we need to do is check $L$ is closed under scalar multiplication. Let $a \in R$ (the ring our module is over).

We know that $g(a\cdot x) = a\cdot g(x)$ (the LHS is the ring action of $M$, the RHS is the ring action of $M/N$), because $g$ is a module homomorphism.

Now if $x \in L$, then $g(x) \in K$, and since $K$ is a submodule of $M/N$ it IS closed under the ring action.

So if $x \in L$, then $a\cdot g(x) \in K$, for any $a \in R$.

But as we saw above, this is $g(a\cdot x)$, so we see that $a \cdot x \in L$, whenever $x \in L$. Thus $L$ is a bona-fide submodule of $M$. But now:

$L/N = g(L) = g(g^{-1}(K)) = K$, so indeed any submodule of $M/N$ is of the form indicated above (since $K$ was chosen arbitrarily).

Is the submodule $L$ we describe above unique? Well, in general, no. But if we insist that $L$ contain $N$, then yes.

Suppose that $g(L) = g(L') = L/N$, and that $N \subseteq L,L'$.

Pick any $x' \in L'$. Then $g(x') = x' + N = l + N$, for some $l \in L$. It follows that $x' - l \in N$, so:

$g(x' - l) = N = 0 + N = 0_{M/N}$.

Since $g$ is a homomorphism, $g(x') = g(l)$. Recall how we defined $L$, as the pre-image under $g$ of $L/N$.

Since $x'$ is such a pre-image, $x' \in L$, and thus $L' \subseteq L$.

A similar argument with the roles of $L,L'$ reversed show that $L \subseteq L'$.

THIS is what mathbalarka means when he says $g^{-1}$ is injective. Given a composition series for $M/N$, we can "pull it back" to a "partial composition series" starting with $N$ and ending with $M$.

We just MERGE that with the composition series for $N$, to obtain a composition series for $M$.

*****************************

Here is an example:

Take the abelian group $\Bbb Z_{210}$. Let's find a composition series for it two different ways:

One such series is:

$\{0\} \subset \langle [105]_{210}\rangle \subset \langle [70]_{210},[105]_{210}\rangle \subset \langle [42]_{210},[70]_{210},[105]_{210}\rangle \subset \Bbb Z_{210}$

For example, we see that $\langle [70]_{210},[105]_{210}\rangle/\langle [105]_{210}\rangle$ has order 3 and so has no non-trivial submodules (in case you're wondering, the cosets are, explicitly:

$[0]_{210} + \langle [105]_{210}\rangle = \{[0]_{210},[105]_{210}\}$
$[35]_{210} + \langle [105]_{210}\rangle = \{[35]_{210},[140]_{210}\}$
$[70]_{210} + \langle [105]_{210}\rangle = \{[70]_{210},[175]_{210}\}$).

This composition series has length 4.

Now let's consider what we get choosing $N = \langle [70]_{210},[105]_{210}\rangle$.

A composition series for $N$ is clearly:

$\{[0]_{210}\} \subset \langle [105]_{210}\rangle \subset N$.

Instead of working with $\Bbb Z_{210}/N$ directly, let's work with the isomorphic group $\Bbb Z_{35}$, obtained from the surjective homomorphism:

$\phi: [k]_{210} \mapsto [k]_{35}$ (Verify that $N$ is indeed the kernel. $N$ is clearly contained in the kernel).

Now $\Bbb Z_{35}/\langle [7]_{35}\rangle$ has order 7, so we obtain the composition series for $\Bbb Z_{35}$ of:

$\{[0]_{35}\} \subset \langle [7]_{35}\rangle \subset \Bbb Z_{35}$.

"Pulling this back" to $\Bbb Z_{210}$, we have:

$\phi^{-1}(\{[0]_{35}\}) = N$
$\phi^{-1}(\Bbb Z_{35}) = \Bbb Z_{210}$

so the only thing of interest is determining $\phi^{-1}(\langle [7]_{35}\rangle)$. That is, we need to find the pre-images of:

$[0]_{35},[7]_{35},[14]_{35},[21]_{35},[28]_{35}$.

These are:

$[0]_{210},[35]_{210},[70]_{210},[105]_{210},[140]_{210},[175]_{210}$ for $[0]_{35}$
$[7]_{210},[42]_{210},[77]_{210},[112]_{210},[147]_{210},[182]_{210}$ for $[7]_{35}$, and so on (keep going up by 7's).

Note that this set of pre-images contains $[42]_{210},[70]_{210},[105]_{210}$, so it contains the submodule:

$\langle [42]_{210},[70]_{210},[105]_{210}\rangle$

Note as well, the submodule of the pre-images has order 30, so it is maximal in $\Bbb Z_{210}$ as 210/30 = 7 is prime. So the pre-image set here actually is the same submodule we used in the first composition series.

These two series have length 2 each, and lo and behold, 2 + 2 = 4.

Thanks for this extensive and informative help, Deveno ... Just working through your post in detail now ...

Thanks to a series of posts by you, Euge and Mathbalarka, I am managing to gain a basic understanding of composition series of modules ...

Peter

 

[Math Amateur]

Something that may, or may not be helpful, here.

When Euge says (in post #9): "...ascends to a series $B_0 \subset B_1 \subset \cdots \subset B_r$.."

he is talking about the same $g^{-1}$ that mathbalarka uses in post #2.

Let's be a bit more explicitly about this "pre-image map" (not an INVERSE map, as $g$ is not injective).

Say we have the map:

$g: M \to M/N$ given by $g(m) = m + N$.

What does a submodule of $M/N$ look like?

I claim it looks like this: $L/N$ for some submodule $L$ of $M$ containing $N$.

It should be clear that if $L$ is a submodule of $M$ containing $N$, that $L/N$ "makes sense", and certainly the set of cosets:

$\{l + N: l \in L\}$ is a submodule of $M/N$.

Just as clearly, this set of cosets ($L/N$) is $g(L)$.

What may NOT be obvious is that ANY submodule of $M/N$ is of this form.

So suppose we have a submodule, say $K$ of $M/N$. Define $L = g^{-1}(K)$.

As far as we know, this is just a set. Suppose $x,y \in L$. Since $g$ is a module homomorphism (and thus an abelian group homomorphism):

$g(x - y) = g(x) - g(y)$.

Since $x,y \in L$, it must be the case that $g(x),g(y) \in K$.

Thus $g(x) - g(y) \in K$ (since this must be true for any two elements of $K$-the "one step group test").

Therefore, $g(x-y) \in L$, so $x - y$ is a pre-image of an element of $K$, and is thus in $L$ (because that is what we defined $L$ to be).

Thus $L$ satisfies the "one step group test", and is thus a subgroup of $(M,+)$. Since $M$ is an abelian group under addition, $L$ is, as well.

So now all we need to do is check $L$ is closed under scalar multiplication. Let $a \in R$ (the ring our module is over).

We know that $g(a\cdot x) = a\cdot g(x)$ (the LHS is the ring action of $M$, the RHS is the ring action of $M/N$), because $g$ is a module homomorphism.

Now if $x \in L$, then $g(x) \in K$, and since $K$ is a submodule of $M/N$ it IS closed under the ring action.

So if $x \in L$, then $a\cdot g(x) \in K$, for any $a \in R$.

But as we saw above, this is $g(a\cdot x)$, so we see that $a \cdot x \in L$, whenever $x \in L$. Thus $L$ is a bona-fide submodule of $M$. But now:

$L/N = g(L) = g(g^{-1}(K)) = K$, so indeed any submodule of $M/N$ is of the form indicated above (since $K$ was chosen arbitrarily).

Is the submodule $L$ we describe above unique? Well, in general, no. But if we insist that $L$ contain $N$, then yes.

Suppose that $g(L) = g(L') = L/N$, and that $N \subseteq L,L'$.

Pick any $x' \in L'$. Then $g(x') = x' + N = l + N$, for some $l \in L$. It follows that $x' - l \in N$, so:

$g(x' - l) = N = 0 + N = 0_{M/N}$.

Since $g$ is a homomorphism, $g(x') = g(l)$. Recall how we defined $L$, as the pre-image under $g$ of $L/N$.

Since $x'$ is such a pre-image, $x' \in L$, and thus $L' \subseteq L$.

A similar argument with the roles of $L,L'$ reversed show that $L \subseteq L'$.

THIS is what mathbalarka means when he says $g^{-1}$ is injective. Given a composition series for $M/N$, we can "pull it back" to a "partial composition series" starting with $N$ and ending with $M$.

We just MERGE that with the composition series for $N$, to obtain a composition series for $M$.

*****************************

Here is an example:

Take the abelian group $\Bbb Z_{210}$. Let's find a composition series for it two different ways:

One such series is:

$\{0\} \subset \langle [105]_{210}\rangle \subset \langle [70]_{210},[105]_{210}\rangle \subset \langle [42]_{210},[70]_{210},[105]_{210}\rangle \subset \Bbb Z_{210}$

For example, we see that $\langle [70]_{210},[105]_{210}\rangle/\langle [105]_{210}\rangle$ has order 3 and so has no non-trivial submodules (in case you're wondering, the cosets are, explicitly:

$[0]_{210} + \langle [105]_{210}\rangle = \{[0]_{210},[105]_{210}\}$
$[35]_{210} + \langle [105]_{210}\rangle = \{[35]_{210},[140]_{210}\}$
$[70]_{210} + \langle [105]_{210}\rangle = \{[70]_{210},[175]_{210}\}$).

This composition series has length 4.

Now let's consider what we get choosing $N = \langle [70]_{210},[105]_{210}\rangle$.

A composition series for $N$ is clearly:

$\{[0]_{210}\} \subset \langle [105]_{210}\rangle \subset N$.

Instead of working with $\Bbb Z_{210}/N$ directly, let's work with the isomorphic group $\Bbb Z_{35}$, obtained from the surjective homomorphism:

$\phi: [k]_{210} \mapsto [k]_{35}$ (Verify that $N$ is indeed the kernel. $N$ is clearly contained in the kernel).

Now $\Bbb Z_{35}/\langle [7]_{35}\rangle$ has order 7, so we obtain the composition series for $\Bbb Z_{35}$ of:

$\{[0]_{35}\} \subset \langle [7]_{35}\rangle \subset \Bbb Z_{35}$.

"Pulling this back" to $\Bbb Z_{210}$, we have:

$\phi^{-1}(\{[0]_{35}\}) = N$
$\phi^{-1}(\Bbb Z_{35}) = \Bbb Z_{210}$

so the only thing of interest is determining $\phi^{-1}(\langle [7]_{35}\rangle)$. That is, we need to find the pre-images of:

$[0]_{35},[7]_{35},[14]_{35},[21]_{35},[28]_{35}$.

These are:

$[0]_{210},[35]_{210},[70]_{210},[105]_{210},[140]_{210},[175]_{210}$ for $[0]_{35}$
$[7]_{210},[42]_{210},[77]_{210},[112]_{210},[147]_{210},[182]_{210}$ for $[7]_{35}$, and so on (keep going up by 7's).

Note that this set of pre-images contains $[42]_{210},[70]_{210},[105]_{210}$, so it contains the submodule:

$\langle [42]_{210},[70]_{210},[105]_{210}\rangle$

Note as well, the submodule of the pre-images has order 30, so it is maximal in $\Bbb Z_{210}$ as 210/30 = 7 is prime. So the pre-image set here actually is the same submodule we used in the first composition series.

These two series have length 2 each, and lo and behold, 2 + 2 = 4.

Hi Deveno ... thanks again for the help ... just checking something that looks to be a typo ... but checking anyway, to be sure ...

You write:

" ... ... As far as we know, this is just a set. Suppose $x,y \in L$. Since $g$ is a module homomorphism (and thus an abelian group homomorphism):

$g(x - y) = g(x) - g(y)$.

Since $x,y \in L$, it must be the case that $g(x),g(y) \in K$.

Thus $g(x) - g(y) \in K$ (since this must be true for any two elements of $K$-the "one step group test").

Therefore, $g(x-y) \in L$, so $x - y$ is a pre-image of an element of $K$, and is thus in $L$ (because that is what we defined $L$ to be). ... ... "In the above argument, you state that

" ... ... $g(x) - g(y) \in K$ ... ..."

and then that

" ... ... Therefore, $g(x-y) \in L$,... ... "

I suspect that it should read " ... ... Therefore, $g(x-y) \in K$,... ... "

Is it a typo as I suspect ... or am I misunderstanding something?

Peter

 

[Math Amateur]

Something that may, or may not be helpful, here.

When Euge says (in post #9): "...ascends to a series $B_0 \subset B_1 \subset \cdots \subset B_r$.."

he is talking about the same $g^{-1}$ that mathbalarka uses in post #2.

Let's be a bit more explicitly about this "pre-image map" (not an INVERSE map, as $g$ is not injective).

Say we have the map:

$g: M \to M/N$ given by $g(m) = m + N$.

What does a submodule of $M/N$ look like?

I claim it looks like this: $L/N$ for some submodule $L$ of $M$ containing $N$.

It should be clear that if $L$ is a submodule of $M$ containing $N$, that $L/N$ "makes sense", and certainly the set of cosets:

$\{l + N: l \in L\}$ is a submodule of $M/N$.

Just as clearly, this set of cosets ($L/N$) is $g(L)$.

What may NOT be obvious is that ANY submodule of $M/N$ is of this form.

So suppose we have a submodule, say $K$ of $M/N$. Define $L = g^{-1}(K)$.

As far as we know, this is just a set. Suppose $x,y \in L$. Since $g$ is a module homomorphism (and thus an abelian group homomorphism):

$g(x - y) = g(x) - g(y)$.

Since $x,y \in L$, it must be the case that $g(x),g(y) \in K$.

Thus $g(x) - g(y) \in K$ (since this must be true for any two elements of $K$-the "one step group test").

Therefore, $g(x-y) \in L$, so $x - y$ is a pre-image of an element of $K$, and is thus in $L$ (because that is what we defined $L$ to be).

Thus $L$ satisfies the "one step group test", and is thus a subgroup of $(M,+)$. Since $M$ is an abelian group under addition, $L$ is, as well.

So now all we need to do is check $L$ is closed under scalar multiplication. Let $a \in R$ (the ring our module is over).

We know that $g(a\cdot x) = a\cdot g(x)$ (the LHS is the ring action of $M$, the RHS is the ring action of $M/N$), because $g$ is a module homomorphism.

Now if $x \in L$, then $g(x) \in K$, and since $K$ is a submodule of $M/N$ it IS closed under the ring action.

So if $x \in L$, then $a\cdot g(x) \in K$, for any $a \in R$.

But as we saw above, this is $g(a\cdot x)$, so we see that $a \cdot x \in L$, whenever $x \in L$. Thus $L$ is a bona-fide submodule of $M$. But now:

$L/N = g(L) = g(g^{-1}(K)) = K$, so indeed any submodule of $M/N$ is of the form indicated above (since $K$ was chosen arbitrarily).

Is the submodule $L$ we describe above unique? Well, in general, no. But if we insist that $L$ contain $N$, then yes.

Suppose that $g(L) = g(L') = L/N$, and that $N \subseteq L,L'$.

Pick any $x' \in L'$. Then $g(x') = x' + N = l + N$, for some $l \in L$. It follows that $x' - l \in N$, so:

$g(x' - l) = N = 0 + N = 0_{M/N}$.

Since $g$ is a homomorphism, $g(x') = g(l)$. Recall how we defined $L$, as the pre-image under $g$ of $L/N$.

Since $x'$ is such a pre-image, $x' \in L$, and thus $L' \subseteq L$.

A similar argument with the roles of $L,L'$ reversed show that $L \subseteq L'$.

THIS is what mathbalarka means when he says $g^{-1}$ is injective. Given a composition series for $M/N$, we can "pull it back" to a "partial composition series" starting with $N$ and ending with $M$.

We just MERGE that with the composition series for $N$, to obtain a composition series for $M$.

*****************************

Here is an example:

Take the abelian group $\Bbb Z_{210}$. Let's find a composition series for it two different ways:

One such series is:

$\{0\} \subset \langle [105]_{210}\rangle \subset \langle [70]_{210},[105]_{210}\rangle \subset \langle [42]_{210},[70]_{210},[105]_{210}\rangle \subset \Bbb Z_{210}$

For example, we see that $\langle [70]_{210},[105]_{210}\rangle/\langle [105]_{210}\rangle$ has order 3 and so has no non-trivial submodules (in case you're wondering, the cosets are, explicitly:

$[0]_{210} + \langle [105]_{210}\rangle = \{[0]_{210},[105]_{210}\}$
$[35]_{210} + \langle [105]_{210}\rangle = \{[35]_{210},[140]_{210}\}$
$[70]_{210} + \langle [105]_{210}\rangle = \{[70]_{210},[175]_{210}\}$).

This composition series has length 4.

Now let's consider what we get choosing $N = \langle [70]_{210},[105]_{210}\rangle$.

A composition series for $N$ is clearly:

$\{[0]_{210}\} \subset \langle [105]_{210}\rangle \subset N$.

Instead of working with $\Bbb Z_{210}/N$ directly, let's work with the isomorphic group $\Bbb Z_{35}$, obtained from the surjective homomorphism:

$\phi: [k]_{210} \mapsto [k]_{35}$ (Verify that $N$ is indeed the kernel. $N$ is clearly contained in the kernel).

Now $\Bbb Z_{35}/\langle [7]_{35}\rangle$ has order 7, so we obtain the composition series for $\Bbb Z_{35}$ of:

$\{[0]_{35}\} \subset \langle [7]_{35}\rangle \subset \Bbb Z_{35}$.

"Pulling this back" to $\Bbb Z_{210}$, we have:

$\phi^{-1}(\{[0]_{35}\}) = N$
$\phi^{-1}(\Bbb Z_{35}) = \Bbb Z_{210}$

so the only thing of interest is determining $\phi^{-1}(\langle [7]_{35}\rangle)$. That is, we need to find the pre-images of:

$[0]_{35},[7]_{35},[14]_{35},[21]_{35},[28]_{35}$.

These are:

$[0]_{210},[35]_{210},[70]_{210},[105]_{210},[140]_{210},[175]_{210}$ for $[0]_{35}$
$[7]_{210},[42]_{210},[77]_{210},[112]_{210},[147]_{210},[182]_{210}$ for $[7]_{35}$, and so on (keep going up by 7's).

Note that this set of pre-images contains $[42]_{210},[70]_{210},[105]_{210}$, so it contains the submodule:

$\langle [42]_{210},[70]_{210},[105]_{210}\rangle$

Note as well, the submodule of the pre-images has order 30, so it is maximal in $\Bbb Z_{210}$ as 210/30 = 7 is prime. So the pre-image set here actually is the same submodule we used in the first composition series.

These two series have length 2 each, and lo and behold, 2 + 2 = 4.

Hi Deveno,

Just a further clarification ...

In the above post you write:

" ... ... What may NOT be obvious is that ANY submodule of $M/N$ is of this form.

So suppose we have a submodule, say $K$ of $M/N$. Define $L = g^{-1}(K)$. ... ... "

So you define L by $L = g^{-1}(K)$, and then later write:

" ... ... Is the submodule $L$ we describe above unique? Well, in general, no. But if we insist that $L$ contain $N$, then yes. ... ... "I am puzzled by your remarks about the uniqueness of L ...

Surely \(\displaystyle g^{-1}(K)\) gives a unique set in \(\displaystyle M\) ... how could the inverse image of a set in \(\displaystyle M/N\) give two (or more) sets in \(\displaystyle M\)? Surely \(\displaystyle g^{-1}(K)\) specifies or determines only one(unique) set in M?

Can you please clarify?

Peter

***EDIT***

(EDIT-1) SORRY ... had a brain fade ... of course you can have two or more sets as an inverse image ... unless the function concerned is injective ... !

Apologies to MHB members for a rather stupid question ...
(EDIT-2) At the beginning of his post Deveno writes:

"Something that may, or may not be helpful, here ... ... "

Definitely is ... ... Overall, the post by Deveno is extremely clear and very helpful ...

 

[Deveno]

Yes, the first was a typo, I have amended it.

As for your second question, while it IS true, $g^{-1}(K)$ defines a unique set, it does not necessarily define a unique set $A$ such that $g(A) = K$, other sets may also possesses this property, and some of them may be submodules of $M$.

So what we are trying to show is that $L$ is the ONLY submodule of $M$ containing $L$ that $g$ maps onto $K$.

 

[Math Amateur]

Yes, the first was a typo, I have amended it.

As for your second question, while it IS true, $g^{-1}(K)$ defines a unique set, it does not necessarily define a unique set $A$ such that $g(A) = K$, other sets may also possesses this property, and some of them may be submodules of $M$.

So what we are trying to show is that $L$ is the ONLY submodule of $M$ containing $L$ that $g$ maps onto $K$.

Just reflecting further on the notion of a pre-image or inverse image ... and I remain a bit confused over the uniqueness of the pre-image ...

I will try to explain my confusion ... and hope you can help to clarify the situation ...Now ...

... if \(\displaystyle f\) is a many-to-one function \(\displaystyle f: A \to B\), and \(\displaystyle B_0\) is a proper, nontrivial subset of \(\displaystyle B\) ...

... and we ask the question:

What subsets \(\displaystyle A_i\) of \(\displaystyle A\) exist such that \(\displaystyle f(A_i) = B_0\)?

... then we get the answer that there may be many of them ...

So, if we define a pre-image of \(\displaystyle B_0\) as any set \(\displaystyle A_i\) such that \(\displaystyle f(A_i) = B_0\) then there may be a number of pre-images of \(\displaystyle B_0\)
On the other hand, as I think is the case, if we define a pre-image of \(\displaystyle B_0\) as the union of all points in \(\displaystyle A\) that map into \(\displaystyle B_0\), then, it appears to me there is only one (unique) pre-image.

Can you please clarify?

Peter
***EDIT***

Just adding an example to make my problem more clear to MHB readers:

Consider the function \(\displaystyle f: A \to B\) and the proper, nontrivial subset \(\displaystyle B_0\), as shown in Figure 1 below:View attachment 3331If the pre-image of \(\displaystyle B_0\) under f is defined as any set\(\displaystyle A_i\) such that \(\displaystyle f(A_i) = B_0\) ... ... then \(\displaystyle A_1\) and \(\displaystyle A_2\) are both pre-images of \(\displaystyle B_0\) ... ...

If the pre-image of \(\displaystyle B_0\) under \(\displaystyle f\) is taken to be \(\displaystyle \{ a \ | \ f(a) \in B_0 \} \) ... ... then the pre-image of \(\displaystyle B_0\) is unique and equal to \(\displaystyle \{ a, b, c, x, y \}\)

I think the pre-image is actually defined to be the latter of the two possibilities ... ... in which case it appears to be unique ... ...

Peter

 

[Deveno]

There's often confusion about this. Here's why:

A function $f: X \to Y$ also defines a similar function:

$\tilde{f}: \mathscr{P}(X) \to \mathscr{P}(Y)$

defined by:

$\forall A \in \mathscr{P}(X): y \in \tilde{f}(A) \iff \exists x \in A: y = f(x)$

the set $\tilde{f}(A)$ is often denoted $f(A)$, but in truth $f$ and $\tilde{f}$ are not the same (they have different domains and co-domains).

This let's us talk about how a function impacts sets, which is often more efficient than talking about how it affects individual elements of sets.

There is a problem with this, however: many functions are not injective, they, in effect, "collapse" the domain to a smaller set. One consequence of this, is that the function $\tilde{f}$ is not a lattice-homomorphism; in particular, it fails to preserve unions.

There is ANOTHER function, however, which IS a lattice-homomorphism:

$g: \mathscr{P}(Y) \to \mathscr{P}(X)$

where:

$\forall B \in \mathscr{P}(Y): x \in g(B) \iff \exists y \in B: f(x) = y$

Note how we have "swapped" the roles of domain and co-domain of $f$ in this definition.

another way to say this is, $g(B)$ is the union of all the sets (and thus union of every singleton) that $f$ maps into $B$.

The function $g$ is typically denoted $f^{-1}$, which can be confusing, since it is usually NOT the same as an inverse function (which may not even exist).

Since $g$ preserves the set-structure (in terms of the boolean lattice induced by set-inclusion), it is the "natural" power-set function to study when investigating SETS, or more generally lattice structures.

Note that $g$ does not give us the COMPLETE structure of $X$, just a sub-lattice (we don't usually recover EVERY subset of $X$). The differences between $\mathscr{P}(X)$ and the sets $\{g(B): B \in \mathscr{P}(Y)\}$ give us an idea of "how faithful" a function $f$ is (how close to being a bijection).

It turns out that if $f$ is "structure-preserving", and we restrict $\tilde{f}$ to certain subsets whose images are singletons (namely: congruence classes), we get an "equivalent structure" to $\tilde{f}(X)$. This is the "seed idea" behind quite a few "isomorphism theorems". We want isomorphisms...we NEED them, and by god, we'll get them, by hook or by crook.

 

[Math Amateur]

There's often confusion about this. Here's why:

A function $f: X \to Y$ also defines a similar function:

$\tilde{f}: \mathscr{P}(X) \to \mathscr{P}(Y)$

defined by:

$\forall A \in \mathscr{P}(X): y \in \tilde{f}(A) \iff \exists x \in A: y = f(x)$

the set $\tilde{f}(A)$ is often denoted $f(A)$, but in truth $f$ and $\tilde{f}$ are not the same (they have different domains and co-domains).

This let's us talk about how a function impacts sets, which is often more efficient than talking about how it affects individual elements of sets.

There is a problem with this, however: many functions are not injective, they, in effect, "collapse" the domain to a smaller set. One consequence of this, is that the function $\tilde{f}$ is not a lattice-homomorphism; in particular, it fails to preserve unions.

There is ANOTHER function, however, which IS a lattice-homomorphism:

$g: \mathscr{P}(Y) \to \mathscr{P}(X)$

where:

$\forall B \in \mathscr{P}(Y): x \in g(B) \iff \exists y \in B: f(x) = y$

Note how we have "swapped" the roles of domain and co-domain of $f$ in this definition.

another way to say this is, $g(B)$ is the union of all the sets (and thus union of every singleton) that $f$ maps into $B$.

The function $g$ is typically denoted $f^{-1}$, which can be confusing, since it is usually NOT the same as an inverse function (which may not even exist).

Since $g$ preserves the set-structure (in terms of the boolean lattice induced by set-inclusion), it is the "natural" power-set function to study when investigating SETS, or more generally lattice structures.

Note that $g$ does not give us the COMPLETE structure of $X$, just a sub-lattice (we don't usually recover EVERY subset of $X$). The differences between $\mathscr{P}(X)$ and the sets $\{g(B): B \in \mathscr{P}(Y)\}$ give us an idea of "how faithful" a function $f$ is (how close to being a bijection).

It turns out that if $f$ is "structure-preserving", and we restrict $\tilde{f}$ to certain subsets whose images are singletons (namely: congruence classes), we get an "equivalent structure" to $\tilde{f}(X)$. This is the "seed idea" behind quite a few "isomorphism theorems". We want isomorphisms...we NEED them, and by god, we'll get them, by hook or by crook.

Thanks for an extremely helpful post, Deveno ...

I note, just for interest, that despite the topic of your post being quite fundamental and important, almost no algebra texts cover this particular issue ... ... So, by the definition of $\tilde{f}$ we could have several sets \(\displaystyle A_i\) such that \(\displaystyle \tilde{f} ( A_i ) = B\) where \(\displaystyle B \subseteq Y\) ( ... but would we call these pre-images of \(\displaystyle B\) under \(\displaystyle f\)? ) ... see an example in Figure 1 below:View attachment 3334In the case of \(\displaystyle g = f^{-1}\), every \(\displaystyle x\) that maps into \(\displaystyle B\) under \(\displaystyle f\) (where \(\displaystyle B \subseteq Y\)) is in \(\displaystyle g( B ) = f^{-1}\) ... ... so, if we use the function \(\displaystyle f\) of Figure 1 and the definition of \(\displaystyle g\) we get Figure 2 below ... and in this case I think \(\displaystyle g(B)\) is the (unique?) pre-image of \(\displaystyle B\) under \(\displaystyle f\) ...View attachment 3335

BUT ... ? ... I am still puzzled over the question of the uniqueness of the pre-image as defined by \(\displaystyle g(B)\) ... it appears to me to be unique by definition ... BUT THEN ... in our previous situation regarding the Lattice Isomorphism Theorem, \(\displaystyle L = g^{-1} (K) \)would be unique by definition and so we would not have to prove its uniqueness ... ... ?

Can you please clarify this situation ... would certainly appreciate your help ... ...

Peter

 

[Deveno]

Again, yes the SET $g^{-1}(K)$ is unique.

That does NOT mean it is the unique set which "has some property".

Let me illustrate with an example:

Consider the $\Bbb Z$-module $\Bbb Z_2 \times \Bbb Z_2$ and the $\Bbb Z$-module homomorphism:

$g: \Bbb Z_2 \times \Bbb Z_2 \to \Bbb Z_2$ given by:

$g(a,b) = a$.

This is a $\Bbb Z$-module homomorphism with kernel $\{0\} \times \Bbb Z_2$.

However, if we let $L$ be the submodule $\Bbb Z_2$ itself (this is not a PROPER submodule, but it is a submodule) there is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ not containing the kernel that $g$ maps onto $\Bbb Z_2$, namely:

$N = \{(0,0),(1,1)\}$

So if our property is:

$K$ is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ that $g$ maps onto $\Bbb Z_2$, there is not a unique sub-module of $\Bbb Z_2 \times \Bbb Z_2$ that does this.

If we change our property to:

$K$ is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ that contains $\text{ker }g$ that $g$ maps onto $\Bbb Z_2$, there is only ONE module possessing this property: $\Bbb Z_2 \times \Bbb Z_2$ itself.

To say a set defined by a function $f$ is unique, is merely to say that the definition is a proper one (it really does specify a set, and only one set). To say a set is the unique set which satisfies some property, is entirely a different matter altogether.

To consider a slightly different example, suppose $R = F$, a field, and $M$ is an $R$-module (vector space) of dimension 3, and that $N$ is a subspace of $M$ of dimension 1 (a line "in $F$").

Suppose that $T:M \to M/N$ is the canonical linear projection. Can you show $\ell(M) = 3$ and $\ell(M/N) = 2$?

The point being, given $K = \langle v + N\rangle$ (for $v \not\in N$) there are are LOTS of planes in $M$ that map to $K$ under $T$. But only ONE of these planes will contain $N$, namely the plane $\langle v,n\rangle$ for some generator $n$ of $N$.

 

[Math Amateur]

Again, yes the SET $g^{-1}(K)$ is unique.

That does NOT mean it is the unique set which "has some property".

Let me illustrate with an example:

Consider the $\Bbb Z$-module $\Bbb Z_2 \times \Bbb Z_2$ and the $\Bbb Z$-module homomorphism:

$g: \Bbb Z_2 \times \Bbb Z_2 \to \Bbb Z_2$ given by:

$g(a,b) = a$.

This is a $\Bbb Z$-module homomorphism with kernel $\{0\} \times \Bbb Z_2$.

However, if we let $L$ be the submodule $\Bbb Z_2$ itself (this is not a PROPER submodule, but it is a submodule) there is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ not containing the kernel that $g$ maps onto $\Bbb Z_2$, namely:

$N = \{(0,0),(1,1)\}$

So if our property is:

$K$ is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ that $g$ maps onto $\Bbb Z_2$, there is not a unique sub-module of $\Bbb Z_2 \times \Bbb Z_2$ that does this.

If we change our property to:

$K$ is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ that contains $\text{ker }g$ that $g$ maps onto $\Bbb Z_2$, there is only ONE module possessing this property: $\Bbb Z_2 \times \Bbb Z_2$ itself.

To say a set defined by a function $f$ is unique, is merely to say that the definition is a proper one (it really does specify a set, and only one set). To say a set is the unique set which satisfies some property, is entirely a different matter altogether.

To consider a slightly different example, suppose $R = F$, a field, and $M$ is an $R$-module (vector space) of dimension 3, and that $N$ is a subspace of $M$ of dimension 1 (a line "in $F$").

Suppose that $T:M \to M/N$ is the canonical linear projection. Can you show $\ell(M) = 3$ and $\ell(M/N) = 2$?

The point being, given $K = \langle v + N\rangle$ (for $v \not\in N$) there are are LOTS of planes in $M$ that map to $K$ under $T$. But only ONE of these planes will contain $N$, namely the plane $\langle v,n\rangle$ for some generator $n$ of $N$.

Thanks so much for your further help on this matter, Deveno ...

just working carefully through rough your post now ... ...

Peter

 

[Math Amateur]

Again, yes the SET $g^{-1}(K)$ is unique.

That does NOT mean it is the unique set which "has some property".

Let me illustrate with an example:

Consider the $\Bbb Z$-module $\Bbb Z_2 \times \Bbb Z_2$ and the $\Bbb Z$-module homomorphism:

$g: \Bbb Z_2 \times \Bbb Z_2 \to \Bbb Z_2$ given by:

$g(a,b) = a$.

This is a $\Bbb Z$-module homomorphism with kernel $\{0\} \times \Bbb Z_2$.

However, if we let $L$ be the submodule $\Bbb Z_2$ itself (this is not a PROPER submodule, but it is a submodule) there is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ not containing the kernel that $g$ maps onto $\Bbb Z_2$, namely:

$N = \{(0,0),(1,1)\}$

So if our property is:

$K$ is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ that $g$ maps onto $\Bbb Z_2$, there is not a unique sub-module of $\Bbb Z_2 \times \Bbb Z_2$ that does this.

If we change our property to:

$K$ is a submodule of $\Bbb Z_2 \times \Bbb Z_2$ that contains $\text{ker }g$ that $g$ maps onto $\Bbb Z_2$, there is only ONE module possessing this property: $\Bbb Z_2 \times \Bbb Z_2$ itself.

To say a set defined by a function $f$ is unique, is merely to say that the definition is a proper one (it really does specify a set, and only one set). To say a set is the unique set which satisfies some property, is entirely a different matter altogether.

To consider a slightly different example, suppose $R = F$, a field, and $M$ is an $R$-module (vector space) of dimension 3, and that $N$ is a subspace of $M$ of dimension 1 (a line "in $F$").

Suppose that $T:M \to M/N$ is the canonical linear projection. Can you show $\ell(M) = 3$ and $\ell(M/N) = 2$?

The point being, given $K = \langle v + N\rangle$ (for $v \not\in N$) there are are LOTS of planes in $M$ that map to $K$ under $T$. But only ONE of these planes will contain $N$, namely the plane $\langle v,n\rangle$ for some generator $n$ of $N$.

Hi Deveno ... thanks again for your help ... BUT ... I need some help in tackling your suggested exercise ...

You write:

" ... ... To consider a slightly different example, suppose $R = F$, a field, and $M$ is an $R$-module (vector space) of dimension 3, and that $N$ is a subspace of $M$ of dimension 1 (a line "in $F$").

Suppose that $T:M \to M/N$ is the canonical linear projection. Can you show $\ell(M) = 3$ and $\ell(M/N) = 2$? ... ... "

Now ... ... to show that $\ell(M) = 3$ and $\ell(M/N) = 2$ we need to show that the number of simple factors \(\displaystyle C_i/C-{i-1}\) in any ascending or descending chain of submodules \(\displaystyle C_i\) for \(\displaystyle M\) is \(\displaystyle 3\) and for \(\displaystyle M/N\) is \(\displaystyle 2\) ... ... but some preliminary concerns ... how do we know that \(\displaystyle M\) and \(\displaystyle M/N\) actually have a composition series ... ... a further issue is that if a module does not have a composition series then (?) presumably it does not have a length ... or is \(\displaystyle \ell(M) = 0 \)... ...

But aside from these preliminary concerns, can you help with the proof that $\ell(M) = 3$ and $\ell(M/N) = 2$?

You help, as always, will be appreciated ...

Peter

 

[Deveno]

Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules).

Now $\text{dim}_F(M/N) = \text{dim}_F(M) - \text{dim}_F(N) = 3 - 1 = 2$.

Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$.

Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$?

You may find it helpful to use that fact that $U = L/N$ for some submodule $N \subset L \subset M$, and that:

$(M/N)/(L/N) \cong M/L$.

In particular, they have the same dimension over $F$ (what has to be the dimension of $L$?).

The point of this, is to show that the length of a module is, in some sense, a generalization of "dimension" for vector spaces. Dimension is founded upon BASES, which, of course, may not exist for a given module. Even so, we may be able to determine a module's length, which gives us a way to "compare size".

 

[Math Amateur]

Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules).

Now $\text{dim}_F(M/N) = \text{dim}_F(M) - \text{dim}_F(N) = 3 - 1 = 2$.

Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$.

Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$?

You may find it helpful to use that fact that $U = L/N$ for some submodule $N \subset L \subset M$, and that:

$(M/N)/(L/N) \cong M/L$.

In particular, they have the same dimension over $F$ (what has to be the dimension of $L$?).

The point of this, is to show that the length of a module is, in some sense, a generalization of "dimension" for vector spaces. Dimension is founded upon BASES, which, of course, may not exist for a given module. Even so, we may be able to determine a module's length, which gives us a way to "compare size".

Thanks Deveno ... just reading your post carefully and studying and reflecting on what you have written ...

Peter

 

[Math Amateur]

Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules).

Now $\text{dim}_F(M/N) = \text{dim}_F(M) - \text{dim}_F(N) = 3 - 1 = 2$.

Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$.

Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$?

You may find it helpful to use that fact that $U = L/N$ for some submodule $N \subset L \subset M$, and that:

$(M/N)/(L/N) \cong M/L$.

In particular, they have the same dimension over $F$ (what has to be the dimension of $L$?).

The point of this, is to show that the length of a module is, in some sense, a generalization of "dimension" for vector spaces. Dimension is founded upon BASES, which, of course, may not exist for a given module. Even so, we may be able to determine a module's length, which gives us a way to "compare size".

Thanks for the help, Deveno ... but ... just a simple clarification ...

You write:

" ... ... Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules). ... ... "
Can you please explain exactly why \(\displaystyle N\) is simple (i.e. why 1-dimensional subspaces of a vector space are simple $F$-modules)?
Peter

 

[Math Amateur]

Recall I said that $N$ is 1-dimensional over $F$, that is:

$N = Fn$, for any non-zero $n \in N$.

It is clear, then, that $N$ is simple as an $F$-module, so a composition series (indeed, the ONLY composition series) is:

$\{0\} \subset N$

so that $\ell(N) = 1$ (1-dimensional subspaces of a vector space are simple $F$-modules).

Now $\text{dim}_F(M/N) = \text{dim}_F(M) - \text{dim}_F(N) = 3 - 1 = 2$.

Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$.

Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$?

You may find it helpful to use that fact that $U = L/N$ for some submodule $N \subset L \subset M$, and that:

$(M/N)/(L/N) \cong M/L$.

In particular, they have the same dimension over $F$ (what has to be the dimension of $L$?).

The point of this, is to show that the length of a module is, in some sense, a generalization of "dimension" for vector spaces. Dimension is founded upon BASES, which, of course, may not exist for a given module. Even so, we may be able to determine a module's length, which gives us a way to "compare size".

Thanks Deveno ...

Now you write:

" ... ... Can you show that:

$N/N \subset U \subset M/N$ is a composition series for $M/N$? ... ... "I need some help with this exercise ... especially with the issue of showing that modules are simple (that is that they contain no proper non-trivial submodules) ... ...

Thoughts so far are as follows:

We are asked:

" ... ... Consider $v \not\in N$ ($v$ is in $M$) and the subspace $U = \langle v + N\rangle$ of $M/N$ ... ... "

so then we have that

\(\displaystyle U = \langle v + N\rangle \)

\(\displaystyle = \{ f ( v + N ) \ | \ f \in F \} \)

\(\displaystyle = \{ f v + N ) \ | \ f \in F \} \)
... and then we are asked to show that:

\(\displaystyle \{ 0 \} = N/N \subset U \subset M/N \)

is a composition series for \(\displaystyle M/N\)So we need to show that the composition factors \(\displaystyle C_i / C_{i - 1}\) are simple for

\(\displaystyle C_1 \subset C_2 \subset C_3 \) \(\displaystyle = \{ 0 \} = N/N \subset U \subset M/N\)That is, we have to show that:

(1) \(\displaystyle (U)/(N/N) = U/\{ 0 \} = U\) is simple

and

(2) \(\displaystyle (M/N) / U = (M/N)/ (L/N) \)
BUT ... for (2) above we have that

\(\displaystyle (M/N)/ (L/N) \cong M/L\)

so, essentially, we have to show \(\displaystyle M/L\) is simple
But ... how to show \(\displaystyle U\) and \(\displaystyle M/L\) are simple ? ... ...

Can you please help?Peter

***EDIT***

You asked about the dimension of \(\displaystyle L\) ... ...

Since \(\displaystyle N \subset L \subset M\)

we have that:

\(\displaystyle \text{ dim } N \lt \text{ dim } L \lt \text{ dim } M\)

BUT ... \(\displaystyle \text{ dim } N = 1\) and \(\displaystyle \text{ dim } M = 3\)

So ... \(\displaystyle \text{ dim } L = 2\)

 

[Deveno]

Let's answer the question:

1) Is a one-dimensional subspace $U$ of a vector space $V$ over a field $F$, a simple $F$-module?

Suppose we have, for the sake of argument:

$\{0_V\} \subset W \subseteq U$.

Since $U$ is one-dimensional, any basis for $U$ is of the form $\{u_1\}$.

This means, given any $u \in U$, there is some $\alpha \in F$ with $u = \alpha u_1$, and moreover, $\alpha$ is unique.

For $\{0_V\} \subset W$ (that is, for this containment to be PROPER), we must have $w \neq 0_V \in W$.

Since $W \subseteq U$, we have $w \in U$. So $w = \beta u_1$.

Since $u_1$ is a basis element, $u_1 \neq 0_V$. If $\beta = 0_F$, we have $w = 0$, contradicting our choice of $w$.

Since $F$ is a field, $\beta^{-1} \in F$, and we have, for any $u \in U$:

$u = \alpha u_1 = [\alpha(\beta^{-1}\beta)]u_1 = (\alpha\beta^{-1})(\beta u_1) = (\alpha\beta^{-1})w \in W$.

This shows that ANY non-zero subspace of $U$ is $U$ itself, that is: $U$ is simple, as it has no non-trivial proper subspaces.

Now if $M$ has dimension 3, and $L$ has dimension 2, $M/L$ has dimension 1. From the above, $M/L$ is simple. This means that:

$N/N \subset L/N \subset M/N$

is a composition series for $M/N$, since:

$(M/N)/(L/N) \cong M/L$ (if $M/L$ had a non-trivial proper subspace, so would $(M/N)/(L/N)$).

 

[Math Amateur]

Let's answer the question:

1) Is a one-dimensional subspace $U$ of a vector space $V$ over a field $F$, a simple $F$-module?

Suppose we have, for the sake of argument:

$\{0_V\} \subset W \subseteq U$.

Since $U$ is one-dimensional, any basis for $U$ is of the form $\{u_1\}$.

This means, given any $u \in U$, there is some $\alpha \in F$ with $u = \alpha u_1$, and moreover, $\alpha$ is unique.

For $\{0_V\} \subset W$ (that is, for this containment to be PROPER), we must have $w \neq 0_V \in W$.

Since $W \subseteq U$, we have $w \in U$. So $w = \beta u_1$.

Since $u_1$ is a basis element, $u_1 \neq 0_V$. If $\beta = 0_F$, we have $w = 0$, contradicting our choice of $w$.

Since $F$ is a field, $\beta^{-1} \in F$, and we have, for any $u \in U$:

$u = \alpha u_1 = [\alpha(\beta^{-1}\beta)]u_1 = (\alpha\beta^{-1})(\beta u_1) = (\alpha\beta^{-1})w \in W$.

This shows that ANY non-zero subspace of $U$ is $U$ itself, that is: $U$ is simple, as it has no non-trivial proper subspaces.

Now if $M$ has dimension 3, and $L$ has dimension 2, $M/L$ has dimension 1. From the above, $M/L$ is simple. This means that:

$N/N \subset L/N \subset M/N$

is a composition series for $M/N$, since:

$(M/N)/(L/N) \cong M/L$ (if $M/L$ had a non-trivial proper subspace, so would $(M/N)/(L/N)$).

Thanks for your help Deveno ... just working carefully through your post now ...

Peter